Let $V$ a $\mathbb{K}-$space of finite dimension. Prove for all pair of different vectors $u,v$ ....

Question

Let $V$ a $\mathbb{K}-$space of finite dimension. Prove for all pair of different vectors $u,v$ exists a linear form $f:V\rightarrow\mathbb{K}$ such that $f(u)\not= f(v)$

My work:

Let $ev_v:V^*\rightarrow\mathbb{K}$ defined by $ev_v(f)=f(v)$ with $v\in V $,
Moreover:

Let $\theta:V\rightarrow V^{**}$ the linear function defined by $\theta (v)=ev_v$

Note $\theta$ is an isomorphism of vector spaces.

Then for $u,v\in V$ such that $u\not=v$ then $f(u)\not=f(v)$

What's your opinion, is good the proof?

Answer 1

Your proof is correct, and quite elegant.

Another proof:

If $\{u, v\}$ is linearly independent, extend $\{u, v\}$ to a basis for $V$, $\{u, v, x_3, \ldots, x_n\}$, and define $f : V \to \mathbb{F}$ using this basis as: $$f(\alpha u + \beta v + \gamma_3x_3 + \ldots \gamma_nx_n) = \alpha$$

Then $f(u) = 1 \ne 0 = f(b)$.

If $\{u, v\}$ is linearly dependent, then assume $v = \lambda u$ for some scalar $\lambda \ne 1$. Extend $\{u\}$ to a basis $\{u, x_2, \ldots, x_n\}$ and define $f : V \to \mathbb{F}$ using this basis as: $$f(\alpha u + \gamma_2 x_2 + \gamma_3x_3 + \ldots \gamma_nx_n) = \alpha$$

Then $f(u) = 1 \ne \lambda = f(\lambda u) = f(v)$.